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In an experiment in space, one proton is held fixed and another proton is released from rest a distance of 2.50 \(\mathrm{mm}\) away. (a) What is the initial acceleration of the proton after it is released? (b) Sketch qualitative (no numberst) acceleration-time and velocity-time graphs of the released proton's motion.

Short Answer

Expert verified
(a) Calculate the force and use Newton's second law to find acceleration. (b) Acceleration decreases over time, velocity increases.

Step by step solution

01

Understanding the Problem

We need to find the initial acceleration of a proton released in space due to the electrostatic repulsion from a nearby proton. The distance between them is given as 2.50 mm.
02

Apply Coulomb's Law

First, we apply Coulomb's law to determine the electrostatic force between the two protons. Coulomb's law is given by \[ F = k \frac{|q_1 q_2|}{r^2} \]where \( k = 8.99 \times 10^9 \, \mathrm{N \, m^2/C^2}\) (Coulomb's constant), \( q_1 \) and \( q_2 \) are the charges of the protons (each \( 1.6 \times 10^{-19} \, C \)), and \( r = 2.50 \, \mathrm{mm} = 2.50 \times 10^{-3} \, \mathrm{m} \).
03

Calculate the Force

Plug the values into Coulomb's law: \[ F = (8.99 \times 10^9) \frac{(1.6 \times 10^{-19})^2}{(2.50 \times 10^{-3})^2} \]Calculate this expression to find the force.
04

Compute the Acceleration

To find the acceleration, use Newton's second law: \[ a = \frac{F}{m} \]The mass \( m \) of a proton is \( 1.67 \times 10^{-27} \, \mathrm{kg}\).Substitute the force calculated earlier to find the initial acceleration \( a \).
05

Sketching Acceleration-Time Graph

Qualitatively, the acceleration of the proton will decrease over time as the distance increases, due to the inverse square nature of the force. The graph starts high and slopes downward.
06

Sketching Velocity-Time Graph

Initially, the velocity of the proton is zero, and it increases over time due to the repulsive force. The velocity-time graph will start at the origin and rise upwards.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Coulomb's Law
Coulomb's Law is fundamental in electrostatics. It describes the force between two charged particles. The law states that the electrostatic force between two charges is directly proportional to the product of their magnitudes and inversely proportional to the square of the distance between them. This force can be calculated using the formula \( F = k \frac{|q_1 q_2|}{r^2} \), where:
  • \( F \) is the electrostatic force,
  • \( k = 8.99 \times 10^9 \, \mathrm{N \, m^2/C^2} \) (Coulomb's constant),
  • \( q_1 \) and \( q_2 \) are the charges of the particles,
  • \( r \) is the distance between the charges.
In the context of protons, since they both have the same positive charge, they repel each other with a force calculated using this law. This principle is crucial for understanding the initial motion of particles in electrostatic fields.
Proton motion
Imagine two protons situated in a vacuum. One is stationary, while the other is initially released. The released proton experiences a repulsive force due to its like charge with the stationary proton. This force accelerates the proton away from its initial position.
The motion of the proton is initially dominated by this electrostatic repulsion. As it accelerates, the distance between the protons increases. This greater distance causes the repulsive force, and thus the acceleration, to decrease over time due to the inverse-square nature of Coulomb's Law. It is crucial to understand this behavior as it helps predict how charged particles move in electrostatic fields.
Acceleration-time graph
An acceleration-time graph visualizes how the acceleration of a particle like our proton changes over time. At the very start, right after release, the acceleration is at its peak due to the close proximity of the two protons. This is because the electrostatic force, which causes acceleration, is strongest at shorter distances.
Over time, as the proton moves away, the acceleration decreases. This decrease happens because the force diminishes with increasing distance, following the inverse-square nature of Coulomb's Law. In a qualitative sense, this graph will show a downward slope, starting high and gradually declining as time progresses.
  • Initially steep, reflecting strong acceleration.
  • Gradually slopes downward as distance increases.
  • Never reaches zero, but approaches it as protons move apart significantly.
Understanding this graph helps in grasping how forces affect motion over time.
Velocity-time graph
The velocity-time graph is key to understanding how the speed of the released proton changes. At the time of release, the proton's velocity is zero since it starts from rest. As it gains speed due to the repulsive force, the velocity increases.
Over time, even though the acceleration decreases, the velocity of the proton continues to rise because it keeps getting pushed, albeit more gently as distance increases. This results in a graph that starts at the origin (velocity zero) and rises sharply initially, then continues to rise at a slower pace.
  • Begins at zero, indicating initial rest state.
  • Rises rapidly due to initial high acceleration.
  • Slope becomes less steep over time as acceleration decreases.
This graph visually represents the cumulative impact of continuous force application on particle velocity.

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Most popular questions from this chapter

A Parallel Universe. Imagine a parallel universe in which the electric force has the same properties as in our universe but there is no gravity. In this parallel universe, the sun carries charge \(Q\) , the earth carries charge \(-Q\) , and the electric attraction between them keeps the earth in orbit. The earth in the parallel universe has the same mass, the same orbital radius, and the same orbital period as in our universe. Calculate the value of \(Q\) . (Consult Appendix F as needed)

Two point charges are placed on the \(x\) -axis as follows: Charge \(q_{1}=+4.00 \mathrm{nC}\) is located at \(x=0.200 \mathrm{m},\) and charge \(q_{2}=+5.00 \mathrm{nC}\) is at \(x=-0.300 \mathrm{m}\) . What are the magnitnde and direction of the total force exerted by these two charges on a negative point charge \(q_{3}=-6.00 \mathrm{nC}\) that is placed at the origin?

(a) What must the charge (sign and magnitude) of a \(1.45-\mathrm{g}\) particle be for it to remain stationary when placed in a downward- directed electric field of magnitude 650 \(\mathrm{N} / \mathrm{C} 2\) (b) What is the magnitude of an electric field in which the electric force on a proton is equal in magnitnde to its weight?

An average human weighs about 650 \(\mathrm{N}\) . If two such generic humans each carried 1.0 coulomb of excess charge, one positive and one negative, how far apart would they have to be for the electric attraction between them to equal their 650 - \(\mathrm{N}\) weight?

A point charge \(q_{1}=-4.00 \mathrm{nC}\) is at the point \(x=0.600 \mathrm{m},\) \(y=0.800 \mathrm{m},\) and a second point charge \(q_{2}=+6.00 \mathrm{nC}\) is at the point \(x=0.600 \mathrm{m}, y=0 .\) Calculate the magnitude and direction of the net electric field at the origin due to these two point charges.

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